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Number of forests of rooted trees with total weight n, where a node at height k has weight 2^k (with root considered to be at height 0).
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%I #39 Sep 25 2023 15:02:56

%S 1,1,1,2,2,3,4,6,7,10,13,17,22,29,38,50,64,82,107,136,175,224,288,363,

%T 465,587,748,942,1196,1503,1902,2385,3004,3765,4729,5911,7406,9246,

%U 11549,14395,17941,22326,27767,34501,42821,53134,65828,81546,100871,124783

%N Number of forests of rooted trees with total weight n, where a node at height k has weight 2^k (with root considered to be at height 0).

%C The sequence b(2n)=0, b(2n+1)=a(n) is the number of trees of weight n.

%H Alois P. Heinz, <a href="/A115593/b115593.txt">Table of n, a(n) for n = 0..5000</a>

%H Hartosh Singh Bal, <a href="https://arxiv.org/abs/2308.00455">Combinatorics of a Class of Completely Additive Arithmetic Functions</a>, arXiv:2308.00455 [math.CO], 2023.

%F Euler transform of b(n), where b(2n+1) = a(n) and b(2n) = 0.

%F From _Paul D. Hanna_, Oct 26 2011: (Start)

%F G.f. satisfies: A(x) = exp( Sum_{n>=1} A(x^(2*n)) * x^n/n ).

%F G.f. satisfies: A(x)*A(-x) = A(x^2). (End)

%F Let b(n) = a(n-1) for n>=1, then sum(n>=1, b(n)*x^n ) = x / prod(n>=1, (1-x^(2*n-1))^b(n) ); compare to A000081, A004111, and A073075. - _Joerg Arndt_, Mar 04 2015

%F a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} ( Sum_{d|k and d is odd} d * a(floor(d/2)) ) * a(n-k). - _Seiichi Manyama_, May 31 2023

%e a(3) = 2; one forest with 3 single-node trees and one with a single two-node tree (root node has weight 1, other node has weight 2).

%p with(numtheory):

%p b:= proc(n) local r; `if`(irem(n, 2, 'r')=0, 0, a(r)) end:

%p a:= proc(n) option remember; `if`(n=0, 1,

%p add(add(d*b(d), d=divisors(j))*a(n-j), j=1..n)/n)

%p end:

%p seq(a(n), n=0..50); # _Alois P. Heinz_, May 16 2013

%t b[n_] := b[n] = If[{q, r} = QuotientRemainder[n, 2]; r == 0, 0, a[q]]; a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d*b[d], {d, Divisors[j]}]*a[n-j], {j, 1, n}]/n]; Table[a[n], {n, 0, 50}] (* _Jean-François Alcover_, Mar 23 2015, after _Alois P. Heinz_ *)

%o (PARI) {a(n)=my(A=1+x); for(i=1, n, A=exp(sum(m=1, n, subst(A,x,x^(2*m)+x*O(x^n))*x^m/m))); polcoeff(A,n)} /* _Paul D. Hanna_ */

%Y Cf. A000081, A117356, A363336, A363337.

%K easy,nonn

%O 0,4

%A _Franklin T. Adams-Watters_, Mar 09 2006